Stresses on 20 degree plane; principal and max shear

Reading of the figure: $\sigma_x = 80$ MPa on the vertical faces (horizontal direction), $\sigma_y = 50$ MPa on the horizontal faces (vertical direction), $\tau_{xy} = 40$ MPa; the plane is at $20^\circ$ to the vertical, so its normal is $20^\circ$ from the $x$-axis.

Stresses on the inclined plane ($\theta = 20^\circ$)

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 65.00 + (15.00)\cos 40^\circ + (40.00)\sin 40^\circ = 102.20\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(15.00)\sin 40^\circ + (40.00)\cos 40^\circ = 21.00\ \text{MPa} \end{aligned}$$

Resultant stress on the plane $= \sqrt{102.20^2+21.00^2} = 104.34\ \text{MPa}$, inclined at $11.61^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 65.00 \pm \sqrt{(15.00)^2+(40.00)^2} = 65.00 \pm 42.72$$

$\sigma_1 = 107.72\ \text{MPa}$ and $\sigma_2 = 22.28\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(40.00)}{30.00} \Rightarrow 2\theta_p = 69.44^\circ$$

So $\theta_p = 34.72^\circ$ (normal of the plane carrying $\sigma_1$) and $124.72^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 42.72\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 79.72^\circ$ and $169.72^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 65.00\ \text{MPa}$.

Check with Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(80, 40)$ and $Y(50, -40)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (65.00, 0)$.
• Draw the circle with centre C and radius $CX = 42.72$.
• It cuts the $\sigma$-axis at $\sigma_1 = 107.72$ and $\sigma_2 = 22.28$; $2\theta_p = \angle XCP_1 = 69.44^\circ$ (scaled by protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm42.72$ with $\sigma = 65.00$.
• Turn CX through $2\theta = 40^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 102.20$ and $\tau \approx 21.00$, which match the analytical values.

Answer: $\sigma_n = 102.20$ MPa, $\tau = 21.00$ MPa; $\sigma_1 = 107.72$ MPa, $\sigma_2 = 22.28$ MPa at $\theta_p = 34.72^\circ$ and $124.72^\circ$; $\tau_{max} = 42.72$ MPa.